A 3.5 g lead bullet moving at 292 m/s strikes a steel plate and stops. If all its kinetic energy is converted to thermal energy and none leaves the bullet, what is its temperature change? Assume the specific heat of lead is 128 J/kg · ◦ C. Answer in units of ◦C.

The change in temperature of the lead bullet is 333.06 ⁰C

Explanation:

Given;

mass of the  lead bullet , m = 3.5 g = 0.0035 kg

velocity of the bullet, v = 292 m/s

specific heat capacity of the lead bullet, c = 128 J/kg.◦ C

let Δθ = change is temperature

Kinetic energy of the bullet is calculated using the formula below;

K.E = ¹/₂mv²

K.E = ¹/₂(0.0035)(292)²

K.E = 149.212 J

If all the kinetic energy of the lead bullet is converted to thermal energy and none leaves the bullet, then the temperature change is calculated using the formula below;

K.E = Q

where;

Q is the quantity of heat generated by the moving bullet or say heat capacity of the lead bullet

Q = mcΔθ

where;

m is mass of the bullet

c is specific heat capacity of the lead bullet

Δθ is the change in temperature

Δθ

Therefore, the change in temperature of the lead bullet is 333.06 ⁰C

 333.06 °C

Explanation:

From the question, and applying the law of conservation of energy

Kinetic energy of the lead bullet = Thermal energy of the lead

1/2mv² = cmΔt Equation 1

Where m = mass of the lead bullet, v = kinetic energy of the lead bullet, c = specific heat capacity of the lead bullet, Δt = change in temperature of the lead bullet.

make Δt the subject of the equation

Δt = 1/2mv²/(cm)

Δt = 1/2v²/c Equation 2

Given: v = 292 m/s, c = 128 J/kg.°C

Δt = 1/2(292²)/128

Δt = 333.06 °C

Hence the change in temperature of the lead bullet =  333.06 °C

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